AMC 10 · 2025 · #4
Grade 8 algebraA team of students is going to compete against a team of teachers in a trivia contest. The total number of students and teachers is 15. Ash, a cousin of one of the students, wants to join the contest. If Ash plays with the students, the average age on that team will increase from 12 to 14. If Ash plays with the teachers, the average age on that team will decrease from 55 to 52. How old is Ash?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A student team and a teacher team together have $15$ people. If a person named Ash joins the students, their average age goes from $12$ to $14$; if Ash instead joins the teachers, their average age goes from $55$ to $52$. Find Ash's age.
Givens: The two teams together have $15$ people: some students and some teachers.; The students' average age is $12$; after Ash joins them it becomes $14$.; The teachers' average age is $55$; after Ash joins them it becomes $52$.; Answer choices: (A) $28$, (B) $29$, (C) $30$, (D) $32$, (E) $33$
Unknowns: Ash's age.; How many students and how many teachers there are (needed along the way).
Understand
Restated: A student team and a teacher team together have $15$ people. If a person named Ash joins the students, their average age goes from $12$ to $14$; if Ash instead joins the teachers, their average age goes from $55$ to $52$. Find Ash's age.
Givens: The two teams together have $15$ people: some students and some teachers.; The students' average age is $12$; after Ash joins them it becomes $14$.; The teachers' average age is $55$; after Ash joins them it becomes $52$.; Answer choices: (A) $28$, (B) $29$, (C) $30$, (D) $32$, (E) $33$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #7 Identify Subproblems
The clues are in words about averages, so Tool #4 (Introduce a Variable) names the unknowns — team sizes $S$, $T$ and Ash's age $A$ — so every sentence can become an equation. Tool #13 (Convert to Algebra) turns each 'new average' condition into a linear equation using total $=$ average $\times$ count. Tool #7 (Identify Subproblems) keeps the student side and the teacher side separate until each has yielded its own equation, after which the two equations plus the total are solved together.
Execute — Answer: A
6.EE.B.6 Step 1 Name the unknowns
- Let $S$ be the number of students, $T$ the number of teachers, and $A$ Ash's age.
- The two teams have $15$ people in all, so the counts add to $15$.
- Naming these three quantities lets each clue in the problem turn into an equation.
💡 Give each unknown a letter so the sentences can turn into equations.
6.SP.B.5 Step 2 Turn the students' clue into an equation
- The $S$ students have a combined age of $12S$, because total age is the average times the count.
- When Ash joins, the team has $S+1$ people and the average becomes $14$, so $\dfrac{12S+A}{S+1}=14$.
- Multiply both sides by $S+1$: $12S+A=14S+14$, which rearranges to $A=2S+14$.
💡 A total is just the average multiplied by how many people share it.
6.SP.B.5 Step 3 Turn the teachers' clue into an equation
- The $T$ teachers have a combined age of $55T$.
- When Ash joins, the team has $T+1$ people and the average drops to $52$, so $\dfrac{55T+A}{T+1}=52$.
- Multiply both sides by $T+1$: $55T+A=52T+52$, which rearranges to $A=52-3T$.
💡 The same total-equals-average-times-count idea works no matter which team Ash joins.
8.EE.C.8 Step 4 Solve the system for the team sizes
- Both expressions equal $A$, so set them equal: $2S+14=52-3T$, which tidies up to $2S+3T=38$.
- Together with $S+T=15$ this is a pair of linear equations.
- From the total, $S=15-T$; substitute: $2(15-T)+3T=38$, so $30+T=38$ and $T=8$.
- Then $S=15-8=7$.
💡 Two straight-line facts about two unknowns pin down exactly one pair of values.
7.EE.B.4 Step 5 Find Ash's age
- Put $S=7$ into $A=2S+14$: $A=2(7)+14=28$.
- As a check, the teachers' equation gives $A=52-3(8)=52-24=28$ — the two agree.
- So Ash is $28$ years old, which is choice (A).
💡 Once the team sizes are known, either age equation hands you Ash's age.
6.EE.B.6 Let $S$ be the number of students, $T$ the number of teachers, and $A$ Ash's age 6.SP.B.5 The $S$ students have a combined age of $12S$, because total age is the average 6.SP.B.5 The $T$ teachers have a combined age of $55T$. When Ash joins, the team has $T+1 8.EE.C.8 Both expressions equal $A$, so set them equal: $2S+14=52-3T$, which tidies up to 7.EE.B.4 Put $S=7$ into $A=2S+14$: $A=2(7)+14=28$. As a check, the teachers' equation giv Review
Reasonableness: An age of $28$ sits well above the students (average $12$) and well below the teachers (average $55$), so it makes sense that Ash pulls the young team's average up and the older team's average down. Both age equations independently return $28$, a strong self-check. The near-miss choices $29$ through $33$ are there to catch a slip such as putting $S$ or $T$ in the denominator instead of $S+1$ or $T+1$, which would produce a different system and a wrong age.
Alternative: Reason directly from how far each average shifts. Ash raises the students' average by $2$ across the new group of $S+1$ people, so $A=12+2(S+1)$. Ash lowers the teachers' average by $3$ across $T+1$ people, so $A=55-3(T+1)$. These simplify to the same two lines as before and give $A=28$ without ever writing a fraction.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming $S$, $T$, and $A$ for the team sizes and Ash's age and writing $S+T=15$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Using average $=$ total age $\div$ count to turn each 'new average' clue into a linear equation.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving $2S+3T=38$ together with $S+T=15$ to get $S=7$, $T=8$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Substituting the team sizes back into $A=2S+14$ to find $A=28$.)
⭐ Turn each 'new average' clue into 'total $=$ average $\times$ count,' then solve the two equations together.
⭐ Turn each 'new average' clue into 'total $=$ average $\times$ count,' then solve the two equations together.
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